Question

Factor the expression completely over the complex numbers.

x^4−625


Factor the expression completely over the complex numbers.

x^4+10x^2+25

Answer 1

`x^4-625=(x^2-25)(x^2+25)=(x-5)(x+5)(x-5i)(x+5i)\\\\x^4+10x^2+25=(x^2+5)^2=((x-\sqrt5i)(x+\sqrt5i))^2=(x-\sqrt5i)^2(x+\sqrt5i)^2`

Answer 2

1)
`x^4-625=(x+5i)(x-5i)(x+5)(x-5)`

2)
`x^4+10x^2+25=(x+√(5)i)^2(x-\sqrt5 i)^2`

Explanation:

1) Given : Expression
`x^4-625`

To find : Factor the expression completely over the complex numbers ?

Solution :

We can re-write the expression as,

`x^4-625=(x^2)^2-(25)^2`

Applying identity,
`a^2-b^2 = (a+b)(a-b)`

`x^4-625=(x^2+25)(x^2-25)`

`x^4-625=(x^2+25)(x^2-5^2)`

Again apply same identity,

`x^4-625=(x^2+25)(x+5)(x-5)`

The factor of
`x^2+25=(x+5i)(x-5i)`

Factor form is
`x^4-625=(x+5i)(x-5i)(x+5)(x-5)`

2) Given : Expression
`x^4+10x^2+25`

To find : Factor the expression completely over the complex numbers ?

Solution :

Expression
`x^4+10x^2+25`

Let
`x^2=y`

`y^2+10y+25`

To factor we equate it to zero.

`y^2+10y+25=0`

Apply middle term split,

`y^2+5y+5y+25=0`

`y(y+5)+5(y+5)=0`

`(y+5)(y+5)=0`

Substitute back,

`(x^2+5)(x^2+5)=0`

`(x^2+5)^2=0`

`x^2+5=0`

`x^2=-5`

Taking root both side,

`x=√(-5)`

`x=\pm √(5)i`

So, The factors are
`(x+√(5)i)(x-\sqrt5 i)`

Factor form is
`x^4+10x^2+25=(x+√(5)i)^2(x-\sqrt5 i)^2`